A linear map $\varphi$ such that $\varphi (GL_n(\mathbb C) )\subseteq GL_n(\mathbb C)$ preserves the rank

Solution 1:

(This is not an answer, but a long comment.)

This examination question is an example of what we call in linear algebra literature a linear preserver problem. I am not going to answer the question in another way, but would like to point out that the solution approach outlined in the examination paper has actually demonstrated several tricks that are rather common in solving linear preserver problems.

Of all these tricks, the most important one is the following:

Rather than considering how a matrix is transformed, consider how a set of matrices is transformed. Characterise each matrix of interest as members of some sets that enjoy some nice properties (geometric, algebraic, topological etc.) properties, and show that these sets are preserved by the linear operator.

The exact matrix sets we look at are case-dependent. Sometimes they are matrix subspaces, sometimes orbits of group actions, sometimes tangent spaces, sometimes extreme points of convex sets and sometimes even intersections of different sets of matrices. In your case, the examination paper considers two kinds of matrix pencils:

1. Matrix pencils of the form $\{M+\lambda A:\ \lambda\in\mathbb C\}$ where $M+\lambda A$ is always nonsingular. Such a set exists if and only if $A$ is singular. If $f$ preserves nonsingular matrices, $f$ also preserves this kind of sets. It follows that $f$ preserves also singular matrices.
2. Matrix pencils of the form $\{N+\lambda A:\ \lambda\in\mathbb C\}$ where $A$ is singular and $N+\lambda A$ is singular for exactly $r$ distinct values of $\lambda$. Such a matrix pencil exists if and and only if $\operatorname{rank}(A)\ge r$. As $f$ preserves both singular and nonsingular matrices, it must also preserve this kind of sets. Hence we get $\operatorname{rank}(f(A))\ge\operatorname{rank}(A)$.

The conclusion of (1) illustrates another trick:

If $f$ preserves a certain set of matrices, it is often useful (but not always possible) to prove that $f$ also preserves the set's complement.

In your case, given that $f$ preserves nonsingular matrices, we also prove that $f$ preserves singular matrices. The latter property is then employed to establish (2).

While the examination paper does not spell out the remaining details, what comes after (2) is another useful trick:

If possible, it is often useful to prove that the linear preserver is bijective and its inverse also preserves the desired property in question.

In your case, having (2), we immediately get that $f$ is bijective. Since $f$ preserves both singular and nonsingular matrices, its inverse $f^{-1}$ must also preserve both singular and nonsingular matrices. So, by applying (2) to $f^{-1}$, we get also $\operatorname{rank}(f^{-1}(A))\ge\operatorname{rank}(A)$. Now the conclusion follows from the two rank inequalities.

Admittedly, the three aforementioned ways of thinking are not always applicable, but more often than not, they are applied successfully in the literature.